Telecommunication system

ABSTRACT

Telecommunication system in which local terminals may be called from one or more central stations by way of transmission paths selected by selection device on the probability of success. Selection takes place with the help of selection tables on the basis of exchanged signalling messages obtained from, or by way of, a switching system (3); other selection tables are also possible, e.g., on the basis of geographic or cartographic information. To the transmission paths there are assigned scores on the basis of the points in time of successful and unsuccessful calls, and of numbers of successful calls. During a session, a call is first made by way of the transmission path having the highest score (S(ok), S(nok) or S(rc)) in one of the selection tables (OK, NOK and RC, respectively). In the event of failure, the call is repeated by way of a preferred transmission path (PP), once again in sequence of the scores from one of the selection tables. If these calls are unsuccessful as well, the other transmission paths are tested, once again in sequence of the scores (S(ok), S(nok) or S(rc)) from one of the selection tables (OK, NOK and RC, respectively). Transmission paths for which it is a short while ago (e.g., less than 5 minutes) that an unsuccessful call took place, are not used. All this is useful, inter alia, for Inmarsat, GSM, UMTS and UPT.

BACKGROUND OF THE INVENTION

The invention relates to a telecommunication system in which local usersare capable of being called from one or more central stations by way oftransmission paths selected by selection means.

An example of such telecommunication system is a cellular system, suchas the GSM system or the future Universal Mobile TelecommunicationSystem (UMTS). The transmission paths comprise base stations which eachtake care of one local region.

Another example is the current Inmarsat system, in which there areincluded, in the transmission paths, satellites which each serve oneregion of the earth's surface.

In this type of system, there is a specific uncertainty as to thelocation where the users are, and therefore also as to the transmissionpath which must be chosen to be capable of reaching a user.

In the GSM system, there are routed calls by way of the transmissionpath which runs by way of the local base station where the user was mostrecently registered. To minimise the uncertainty regarding the userlocation, it is necessary that the user call regularly by having hismobile terminal regularly transmit a code. In the event of "pockettelephones", the transmission of such code requires a relatively greatdeal of battery power, as a result of which the battery must berecharged regularly.

In the Inmarsat system, so far calls are routed to the (four) regions onthe basis of the region indicated by the calling party. If a call isunsuccessful, the call is repeated in one, two or three other regions.In practice, all of this turns out to lead to many calls beingunsuccessful.

SUMMARY OF THE INVENTION

The object of the invention is to improve the routing of calls by, onthe basis of the system signalling, which may be derived from thesystem, prior to a call to a local user calculating a selection order,according to the probabilities per transmission path that the call isanswered by the local user. Thereafter, the local user is called, by wayof transmission paths, in descending probability order, until the callis answered.

The invention is suitable for use in transmission systems having mobileterminals, such as Inmarsat, GSM and UMTS, but also for systems, such asthe UPT system, in which a user, sequentially or simultaneously, atvarious terminals, mobile or non-mobile, may choose domicile by havinghimself registered from there at the system server. In the event of sucha system, where the user terminal is logically mobile, the same problemapplies regarding the uncertainty of the user location as in systemswhere terminals are physically mobile.

In accordance with the invention, it is calculated by way of whichtransmission path the local user may best be called, with the highestprobability of success. If a call by the local user is not answered,another transmission path having a lower probability of success ischosen.

The selection order is preferably calculated by means of one or moreselection tables which each comprise a designation of the replyprobabilities per transmission path. Said selection tables are theresult of registrations, in the central stations, of code signalstransmitted to, or received from, the local users. The use of variousselection tables is based on the insight that the best reply probabilityis the resultant of an order of various factors which may be derived(inter alia) from the signalling of the transmission system.

A first selection table is formed, e.g., using the recentness of thepoints in time, registered in the central stations, of code signalswhich were received from the local user by way of the varioustransmission paths. From the user, there is received a (signalling) codesignal when the user has himself registered in the system, or inresponse to a call to said user. In either case, the more recent thereceived code signal, the greater the possibility that the user (still)is in the region which is served by the same transmission path. It istherefore logical, in the event of a new call to the user, to first usethe transmission path by way of which a code signal was most recentlyreceived from the user.

If, after receipt of a code signal from the user, there is made a callwhich is not answered, however, it is logical to assume that the user isno longer in the region which is served by the transmission path. Theobservation that the user does not answer a call by way of a certaintransmission path, minimises the probability that a new call by way ofthat same transmission path will be answered. Said probability, however,grows again as time elapses: if the failure of the call occurred aconsiderable time ago, a new call again has a fair chance of beinganswered. In order therefore to be capable of correcting, by negativeobservations, the positive observations regarding the accessibility ofthe user, which are ranked in the first selection table, there may beformed a second selection table using the recentness of the points intime registered in the central stations of code signals which were sentto the local user by way of the various transmission paths, but which,as shown by the absence of the receipt, within a certain period of time,of a code signal in response thereto, were not answered by the localuser.

The first selection table and the second selection table may be used aseach other's opposites: a positive observation erases the effect of anearlier negative observation, while a negative observation annuls theeffect of an earlier positive observation. In addition, the effect ofeach observation decreases as time elapses; in other words, a recentobservation prevails over a less recent observation. Combination of bothselection tables amplifies the surmise of the region or location wherethe user is and, by selecting said region or location, increases theprobability of a reply.

Still further observations, ranked and laid down in further selectiontables, may improve the selection process.

Thus, a third selection table may be formed using the numbers,registered in the central stations, of code signals per transmissionpath originating from the local user. In this connection, therefore, asa selection criterion there is used the number of times that the userlogged on to the system by way of the various transmission paths, eitherof his own accord or as a result of a call from the system. It islogical to assume that, definitely if further designations are lacking,the probability of a reply to a new call is greatest in a region fromwhich the user has been the most active so far. Incidentally, in thisconnection, too, it is desirable to assign a greater weight to morerecent registrations than to registrations carried out longer ago.

Yet another, fourth selection table may be formed by a table in whichfor each region the adjacent regions are designated. The logic in theuse of said table, as in selecting regions having a good replyprobability, is the assumption that a user⁻⁻ if a call is stillunsuccessful in one region where the user, as is shown by recentobservation, recently was⁻⁻ will then be in an adjacent region.Particularly for large regions, such as in the event of Inmarsat, thisis a sound assumption. An extension⁻⁻ which is particularly useful forsystems having small regions⁻⁻ is to construct a probable path followedby the user through the regions using consecutive observations, it beingpossible to obtain the region in which the user will probably be, byextrapolation. During said extrapolation, cartographic data may also beused: there may be taken account of, e.g., the course of roads and thelike. In general, therefore, after a call by way of the path having thehighest probability score still was unsuccessful, a next call ispreferably made by way of a path which, based on logical considerationsregarding user movements, has the greatest probability of success.Below, in this connection there will be referred to "preferredtransmission paths", with each transmission path having one or morepreferred transmission paths.

The various selection tables, which all offer a certain designation withrespect to the probability of success, may be combined with one anotherin various ways and in various orders. During a session, a call is firstmade by way of the transmission path having the highest score in thefirst, second or third selection table. In the event of failure, thecall may be repeated by way of a preferred transmission path as laiddown in the fourth selection table, in sequence of the scores from thefirst, second or third selection table. Should said calls beunsuccessful as well, the remaining transmission paths are tested, againin sequence of the scores from one of the selection tables. Transmissionpaths for which it is a short while ago (e.g., less than 5 minutes) thatan unsuccessful call took place, are not used. In the next paragraph,several options are discussed.

REFERENCES BRIEF DESCRIPTION OF THE DRAWING

Below, the invention will be explained in greater detail by reference toFIG. 1, in which the link is shown between a fixed terminal and a mobileterminal, partly by way of the fixed network, partly by way of asatellite link. Furthermore, the construction of the selection order ofcalls will be discussed by reference to a number of selection tables.

DETAILED DESCRIPTION

FIG. 1 shows a terminal 1, connected to a telecommunication network 2.By way of an intelligent switching system 3, a call may be made to amobile user terminal 6 by way of earth stations 4 (one of which is shownin the drawing) and a system of (in the event of Inmarsat four)satellites 5. In the switching system 3, there is chosen, using variousdata collected there, the transmission path (earth station-satellite)whose probability that the call is answered by the local user 6, isgreatest. By way of the switching system 3, the signalling messages(code signals) exchanged between the user terminal 6 and the earthstation 4 (by way of the satellites 5) are continuously tapped, andstored in a data base 7. The switching system 3 is capable of receivingand registering and storing the code signals exchanged by all earthstations 6.

In the data base of the switching system 3, per terminal 6, for eachtransmission path separately there is stored:

OK: the point in time on which a code signal was received from themobile user terminal 6 for the last time, whether or not in response toa call from the earth station 4;

NOK: the point in time on which, as shown by the failure of a codesignal forthcoming from the user terminal 6 in response to a call fromthe earth station 4, an unsuccessful call was made to the terminal 6 forthe last time;

RC: the total number of times that, over a certain period of time, acode signal was received from the terminal. To said value, there may beapplied "exponential smoothing", as a result of which the effect ofolder annotations is reduced, to the benefit of more recentregistrations. This may be achieved by periodically dividing the RCvalue by a constant value >1.

Prior to a new call, said variables are used for calculating a regionorder relating to the probability of success of said new call. In saidcalculation, there are also involved the regions which physically adjointhe region in which a successful call took place most recently.

Below, there follow three examples of a calculation of thetransmission-path order for four transmission regions. Subsequently, afourth example will present an identical calculation for a greaternumber of transmission paths. The meaning of the abbreviations used is:

    ______________________________________                                        PN =   path number; an * denotes the path having the highest CS(ok)                  score;                                                                 OK =   the most recent point in time, counted from t = 0, on which                   a code signal was received from the user terminal 6;                   S(ok) =                                                                              score based on OK times;                                               NOK =  the point in time on which a call to the terminal was                         unsuccessful;                                                          S(nok) =                                                                             score based on NOK times;                                              CS(ok) =                                                                             score based on OK times, however corrected using the NOK                      times;                                                                 PP(##) =                                                                             preferred paths, associated with the path having the highest                  RS(ok) score;                                                          RC =   total responses received per path of the terminal;                     S(rc) =                                                                              score based on RC values;                                              CPO =  calculated path order.                                                 ______________________________________                                    

The steps followed are:

1. Rank the paths using the OK times; the more recent the OK times, thehigher the score (S(ok)).

2. Investigate for all paths whether the NOK time is shorter than the OKtime; in that case the score becomes 0 (CS(ok)).

3. Rank the paths using the NOK times; the longer the NOK times, thehigher the score (S(nok)).

4. If after step 2 all scores are equal to 0, the scores (CS(ok)) aremade equal to the scores from step 3; otherwise, the score remains as itwas determined in steps 1 and 2.

5. Determine the preferred paths (PP(##)) of the path having the highestscore.

6. Rank the preferred paths using the numbers of responses received(S(rc)).

7. The calculated path order (CPO) is: the path having the highest scoredetermined in steps 1, 2 and 4, and then the preferred paths, ranked asin step 6.

After the examples below, a similar, yet improved method will bepresented.

EXAMPLE

    ______________________________________                                                OK     NOK                                                            PN      in s.  in s.    RC  S(ok)  S(nok)                                                                              CS(ok)                               ______________________________________                                        0       -446   -276     2   2      0     0                                    *1      -111   -822     81  3      2     3                                    2       -481   -809     34  1      1     1                                    3       -853   -829     24  0      3     0                                    ______________________________________                                        PP(1)   RC     S(rc)                                                          ______________________________________                                        3       24     0                                                              2       34     1                                                              ______________________________________                                         CPO: 1-                                                                      2-3                                                                       

The operation is as follows:

From the data base of the switching system 3 there are derived, perterminal 6, for the transmission paths 0 . . . 3, the points in time onwhich a code signal was received from the user terminal 6 for the lasttime, either because the terminal 6 was registered by way of said path,or as a reaction to a call by way of that path. The shorter the timeelapsed since the terminal emitted a code signal, the greater theprobability that the terminal will be accessible by way of the samechannel; therefore, the shorter the OK time, the higher theaccessibility score shown under S(ok).

From the data base of the switching system 3, there are also derived,per terminal 6, for the transmission paths 0 . . . 3, the points in timeon which an unanswered call was made to the user terminal 6 for the lasttime. The shorter the time elapsed since the terminal evidently was notaccessible, the greater the probability that the terminal will not beaccessible either by way of the same channel; therefore, the shorter theNOK time, the lower the accessibility score, shown under S(nok).

If for a certain path at a certain moment (OK time) there proved to becontact with the terminal, but thereafter the terminal proved to be nolonger accessible (NOK time), the effect of the OK time is annulled bythe (more recent) NOK time. The OK scores are thus corrected: the scoresS(ok) of the paths for which the absolute value of the NOK time is lessthan that of the OK time, are set to 0. In the above example, such isthe case for path 0, there having been contact with the terminal 446seconds ago, resulting in an OK score of 2 with respect to the otherpaths. Since, however, 276 seconds ago, due to an unsuccessful call tothe terminal by way of said path 0, the terminal proved no longeraccessible by way of path 0, the OK score is returned to 0 (see underCS(ok)). It may occur that in this manner all OK scores must becorrected to 0, as is the case in example 3 below. In this case, thescore is determined using the NOK times; after all, the OK times nolonger offer any basis for calculating the greatest accessibilityprobability.

When the terminal 6 is called, the path having the greatest (corrected)score CS(ok)⁻⁻ path 1 in example 1⁻⁻ has the greatest probability ofsuccess. In the event, however, that the call still were to beunsuccessful, the call is repeatedly emitted by way of a path which hasa special relation to path 1. Thus, it is logical to assume that theterminal, after an unsuccessful call by way of the most probable path 1,or by way of the path which ends in an adjacent region, will beaccessible. In example 1, said preferred paths PP(1) of path 1 are thepaths 3 and 2.

The order in which said alternative paths 3 and 2 will be used iscalculated using the total numbers of (successful) links which in thepast have taken place between the switching system 3 and the terminal byway of said paths 3 and 2. Having said this, such order may also bederived from one of the other tables, e.g., the OK table. In the columnRC there is shown, for all paths, the total number of links. From this,there prove to have been 24 links by way of path 3, and 34 links by wayof path 2. It is logical to assume that the probability of connection byway of path 2 is greater than by way of path 3; path 2 therefore scoreshigher than path 3.

The calling sequence (CPO) therefore becomes: path 1⁻⁻ path 2⁻⁻ path 3.

EXAMPLE

    ______________________________________                                                OK     NOK                                                            PN      in s.  in s.    RC  S(ok)  S(nok)                                                                              CS(ok)                               ______________________________________                                        0       -600   -667     57  0      2     0                                    1       -322   -227     37  3      1     0                                    2       -591   -202     1   1      0     0                                    *3      -445   -922     42  2      3     2                                    ______________________________________                                        PP(3)   RC     S(rc)                                                          ______________________________________                                        1       37     1                                                              2       1      0                                                              ______________________________________                                         CPO: 3-                                                                      1-2                                                                       

Example 2 shows the result of the calculation method according to theinvention for other input variables (OK, NOK, PP(##) and RC). In thisexample, the highest OK score⁻⁻ for path 1⁻⁻ is annulled by an NOK time(⁻⁻ 227) which is more recent than the OK time (⁻⁻ 322). As a result, topath 3 there is assigned the highest probability of success for a callto the terminal. The preferred paths of path 3 are the paths 1 and 2, ofwhich path 1 in total has the most switching system/terminal links toits credit (37). The call order therefore becomes path 3⁻⁻ path 1⁻⁻ path2.

EXAMPLE

    ______________________________________                                                OK     NOK                                                            PN      in s.  in s.    RC  S(ok)  S(nok)                                                                              CS(ok)                               ______________________________________                                        0       -852   -305     64  1      2     2                                    *1      -910   -894     18  0      3     3                                    2       -481   -293     57  3      1     1                                    3       -574   -261     51  2      0     0                                    ______________________________________                                        PP(1)   RC     S(rc)                                                          ______________________________________                                        2       57     0                                                              0       64     1                                                              ______________________________________                                         CPO: 1-                                                                      0-2                                                                       

In example 3, the phenomenon occurs that for all four paths the NOKtimes are more recent than the OK times, as a result of which a scorebased on the OK times is not possible. In this case, therefore, use ismade of the NOK times for determining the path having the greatestchance of success. Based on the NOK times, path 1 receives the highestscore (see under S(nok)). Thereafter, the operation again takes place inthe same manner: of path 1, the preferred paths are determined (2 and 0)of which the score is then determined on the basis of the RC values. Theresult is the path order 1-0-2.

Finally, shown below is example 4, generated in the same manner as theprevious examples, applied however to a transmission system in whichbasically links by way of 25 different paths may be set up with the sameterminal 6. In this example, it has been taken into account that eachpath has four preferred paths.

    ______________________________________                                                OK     NOK                                                            PN      in s.  in s.    RC  S(ok)  S(nok)                                                                              CS(ok)                               ______________________________________                                         0      -486   -268     75  13     6     0                                     1      -809   -711     4   4      17    0                                     2      -861   -528     9   2      13    0                                     3      -116   -772     37  23     20    23                                    4      -981   -514     71  1      12    0                                     5      -378   -584     87  16     15    16                                   *6      -8     -296     75  25     8     25                                    7      -691   -916     24  9      24    9                                     8      -447   -923     84  14     25    14                                    9      -221   -335     45  19     9     19                                   10      -409   -627     35  15     16    15                                   11      -592   -35      70  11     1     0                                    12      -128   -269     96  22     7     22                                   13      -524   -912     90  12     23    12                                   14      -723   -50      84  7      2     0                                    15      -729   -897     97  6      22    6                                    16      -218   -462     18  20     10    20                                   17      -47    -529     46  24     14    24                                   18      -808   -488     81  5      11    0                                    19      -841   -787     35  3      21    0                                    20      -368   -57      40  17     3     0                                    21      -633   -165     94  10     5     0                                    22      -227   -749     14  18     19    18                                   23      -996   -3       28  0      0     0                                    24      -178   -747     13  21     18    21                                   25      -698   -64      44  8      4     0                                    ______________________________________                                        PP(6)   RC     S(rc)                                                          ______________________________________                                         2      9      1                                                               1      4      0                                                               3      37     2                                                               5      87     3                                                              ______________________________________                                         CPO: 6-5-3-2-1                                                           

Below, there is given a modified method of calculating the path order.The steps thereof are:

1. Rank the paths using the OK times; the more recent the OK times, thehigher the score (S(ok)).

2. Of the path having the highest score, determine the preferred paths(PP(##)).

3. Rank the preferred paths using the numbers of responses per path(S(rc)).

Remark: The preferred paths might be ranked once again using the OKtimes, or possibly using preferences designated by the callingsubscriber (this remark also applies to the previous method).

4. Rank the remaining, non-preferred paths using the response numbers(or OK times).

5. Determine whether there are recent NOK times (in the examples belowless than 300 s.).

6. The path order then becomes: the path, selected under 1, having thehighest OK score; then ranked preferred paths; and finally thenon-preferred paths, it being understood that paths having recent NOKtimes are ruled out. Should all transmission paths be ruled out as aresult, the entire procedure is repeated from step 1.

Processing the first example using these steps leads to the followingresult.

EXAMPLE

    ______________________________________                                                  OK     NOK                                                          PN        in s.  in s.        RC  S(ok)                                       ______________________________________                                        0         -446   -276         2   2                                           1         -111   -822         81  3                                           2         -481   -809         34  1                                           3         -853   -829         24  0                                           PP(1)     RC     S(rc)                                                        ______________________________________                                        3         24     0                                                            2         34     1                                                            ______________________________________                                         CPO: 1-                                                                      2-3                                                                       

The operation is as follows.

From the steps 1, 2 and 3, there results an order corresponding to theone from example 1: 1-2-3. A step omitted in the previous examples isstep 4, in which the remaining paths are ranked as well. The path orderthen becomes 1-2-3-0. According to step 5, however, path 0 is ruled outagain on account of a recent NOK time <300 s.

Processing the second example in conformity with the second method takesplace as follows:

EXAMPLE

    ______________________________________                                                  OK     NOK                                                          PN        in s.  in s.        RC  S(ok)                                       ______________________________________                                        0         -600   -667         57  0                                           *1        -322   -227         37  3                                           2         -591   -202         1   1                                           3         -445   -922         42  2                                           ______________________________________                                        PP(1)     RC     S(rc)                                                        ______________________________________                                        3         24     0                                                            2         34     1                                                            ______________________________________                                         CPO: 3-                                                                  

From the steps 1, 2 and 3 there follows an order 1-2-3. By means of step4 path 0 is added. The path order then becomes 1-2-3-0. According tostep 5, however, the paths 1 and 2 are ruled out again, on account of arecent NOK time <300 s., so that the order becomes 3-0.

Below, there finally follows yet another example, in which a systemhaving transmission paths 0 . . . 25 calculates the path order accordingto the second method.

EXAMPLE

    ______________________________________                                                OK         NOK                                                        PN      in s.      in s.  RC      S(ok)                                                                              S(rc)                                  ______________________________________                                         0      -232       -167   52      16   12                                      1      -18        -937   4       24   0                                       2      -447       -52    57      9    14                                      3      -20        -205   11      23   2                                       4      -71        -154   76      21   20                                      5      -49        -365   95      22   24                                      6      -97        -801   16      20   5                                       7      -180       -6     65      18   15                                      8      -326       -906   53      14   13                                      9      -738       -963   80      6    22                                     10      -817       -605   42      4    9                                      11      -378       -81    70      13   17                                     12      -590       -147   13      7    3                                      13      -414       -238   96      10   25                                     14      -582       -38    28      8    7                                      15      -392       -602   29      12   8                                      16      -846       -428   7       3    1                                      17      -924       -152   23      1    6                                      *18     -8         -753   44      25   10                                     19      -762       -443   14      5    04                                     20      -968       -628   76      0    21                                     21      -129       -619   70      19   18                                     22      -852       -982   70      2    19                                     23      -296       -683   68      15   16                                     24      -219       -454   46      17   11                                     25      -408       -8     89      11   23                                     ______________________________________                                        PP(18)  RC         S(rc)                                                      ______________________________________                                        11      70         1                                                          25      89         2                                                          14      28         0                                                          13      96         3                                                          ______________________________________                                         CPO': 18 13 25 11 14 5 9 20 4 22 21 23 7 2 8 0 24 10 15 14 17 6 19 12 3 1     1                                                                             CPO": 18 x x x x 5 9 20 x 22 21 23 x x 8 0 24 10 15 14 x 6 19 x x 16 1        CPO: 18 5 9 20 22 21 23 8 0 24 10 15 14 6 19 16 1                        

First, the calculated order (CPO') is shown without taking the NOK timesinto account; subsequently (CPO"), the paths whose NOK times are <300 s.are ruled out from selection ("x"); and finally the resulting order ispresented (CPO).

The various examples for this application were generated with the helpof a GWBASIC program whose source code is shown in Table

                                      TABLE                                       __________________________________________________________________________    100 REM SAVE"PN.BS", A                                                        110 CLS : KEY OFF                                                             120 OK = 1: NOK = 0: MAX.PATH = 50                                            130 DIM TIME(1, MAX.PATH), SCORE(1, MAX.PATH), CORRECTED.SCORE(2,             MAX.PATH), PREFERRED.PATH(MAX.PATH, MAX.PATH),                                RECEIVED.CODES(MAX.PATH), RANKED.SCORE(1, MAX.PATH),                          PREFERRED.PATH$(MAX.PATH, MAX.PATH), MARK$(MAX.PATH),                         PREF.SCORE(MAX.PATH), PREF.PATH$(MAX.PATH), NOPRINT(MAX.PATH)                 140 `---INPUT NUMBER OF PATHS---------------------------150 OPEN "PN.OP"      FOR OUTPUT AS #1: LOCATE 1, 1: PRINT #1, STRING$(70, ""):LAST.PATH(0) =       LAST.PATH: LOCATE 1, 1: INPUT "Paths 0 . . . ", LAST.PATH$: LAST.PATH =       VAL(LAST.PATH$)                                                               160 IF LAST.PATH$ = ""THEN LAST.PATH = LAST.PATH(0)                           170 PREF.PATH = INT(LAST.PATH / 2): IF PREF.PATH > 3 THEN PREF.PATH = 3       180 IF LAST.PATH$ = "0"THEN CLOSE: CLS : END                                  190 `---RESET VARIABLES------------------------------200RANKED.SCORE =        0;                                                                            CORRECTED.SCORES = 0: RANKED.SCORE.MAX = 0                                    210 FOR PATH = 0 TO LAST.PATH                                                 220 RANKED.SCORE(OK, PATH) = 0                                                230 PREF.SCORE(PATH) = 0                                                      240 NEXT PATH                                                                 250 `---GENERATE INPUT DATA--------------------------------                   260 RANDOMIZE TIMER                                                           270 FOR PATH = 0 TO LAST.PATH                                                 280 TIME(OK, PATH) = -INT(RND * 1000) + 1                                     290 TIME(NOK, PATH) = -INT(RND * 1000) + 1                                    300 NEXT PATH                                                                 310 IF LAST.PATH$ = ""GOTO 460                                                320 FOR PATH = 0 TO LAST.PATH                                                 330 RECEIVED.CODES(PATH) = INT(RND * 100) + 1                                 340 NEXT PATH                                                                 350 IF LAST.PATH$ = ""GOTO 460                                                360 FOR PATH = 0 TO LAST.PATH                                                 370 FOR X = 0 TO PREF.PATH                                                    380 PREFERRED.PATH(PATH, X) = INT(RND * (LAST.PATH + 1))                      390 FOR Y = 0 TO PREF.PATH                                                    400 IF (X <> Y) AND (PREFERRED.PATH(PATH, X) = PREFERRED.PATH(PATH, Y))       GOTO 380                                                                      410 IF PREFERRED.PATH(PATH, X) = PATH GOTO 380                                420 NEXT Y                                                                    430 PREFERRED.PATH$(PATH, X) = MID$(STR$(PREFERRED.PATH(PATH, X)), 2):        IF                                                                            (LAST.PATH >= 10) AND (PREFERRED.PATH(PATH, X) < 10) THEN                     PREFERRED.PATH$(PATH, X) = ""+ PREFERRED.PATH$(PATH, X)                       440 NEXT X                                                                    450 NEXT PATH                                                                 460 `---CALCULATE OK-SCORE USING POSITIVE RESPONSE DATA----------470 FOR      PATH = 0 TO LAST.PATH                                                         480 SCORE(OK, PATH) = 0                                                       490 FOR X = 0 TO LAST.PATH                                                    500 IF TIME(OK, PATH) > TIME(OK, (PATH + X) MOD (LAST.PATH + 1)) THEN         SCORE(OK, PATH) = SCORE(OK, PATH) + 1                                         510 NEXT X                                                                    520 NEXT PATH                                                                 530 `GOTO 640                                                                 540 `---CALCULATE NOK-SCORE USING NEGATIVE RESPONSE DATA----------            550 FOR PATH = 0 TO LAST.PATH                                                 560 SCORE(NOK, PATH) = 0                                                      570 FOR X = 0 TO LAST.PATH                                                    580 IF TIME(NOK, PATH) < TIME(NOK, (PATH + X) MOD (LAST.PATH + 1)) THEN       SCORE(NOK, PATH) = SCORE(NOK, PATH) + 1                                       590 NEXT X                                                                    600 NEXT PATH                                                                 610 `---CORRECT OK-SCORE USING NEGATIVE RESPONSE DATA-----------620 FOR       PATH = 0 TO LAST.PATH                                                         630 `IF TIME(OK, PATH) < TIME(NOK, PATH) THEN CORRECTED.SCORE(OK, PATH)       0: ELSE CORRECTED.SCORE(OK, PATH) = SCORE(OK, PATH)                           640 NEXT PATH                                                                 650 `---RE-RANK CORRECTED SCORE AND CALCULATE HIGHEST SCORE-------            660 FOR PATH = 0 TO LAST.PATH                                                 670 `MARK$(PATH) = ""                                                         680 FOR X = 0 TO LAST.PATH                                                    690 CORRECTED.SCORE(OK, PATH) = SCORE(OK, PATH)                               700 IF CORRECTED.SCORE(OK, PATH) > CORRECTED.SCORE(OK, (PATH + X) MOD         (LAST.PATH + 1)) THEN RANKED.SCORE(OK, PATH) = RANKED.SCORE(OK, PATH) +       710 IF CORRECTED.SCORE(OK, PATH) = 0 THEN RANKED.SCORE(OK, PATH) =            RANKED.SCORE: RANKED.SCORE = RANKED.SCORE + 1: GOTO 730                       720 NEXT X                                                                    730 CORRECTED.SCORES = CORRECTED.SCORES + CORRECTED.SCORE(OK, PATH)           740 IF RANKED.SCORE(OK, PATH) > RANKED.SCORE.MAX THEN RANKED.SCORE.MAX        = RANKED.SCORE(OK, PATH)                                                      750 NEXT PATH                                                                 760 FOR PATH = 0 TO LAST.PATH                                                 770 IF CORRECTED.SCORES = 0 THEN CORRECTED.SCORE(OK, PATH) = SCORE(NOK,       PATH): RANKED.SCORE(OK, PATH) = SCORE(NOK, PATH)                              780 `IF RANKED.SCORE(OK, PATH) = LAST.PATH THEN SELECTED.PATH = PATH          790 IF RANKED.SCORE(OK, PATH) = RANKED.SCORE.MAX THEN SELECTED.PATH =         PATH                                                                          800 NEXT PATH                                                                 810 `---RANK FOR NUMBER OF RECEIVED RESPONSES OF PREFERRED PATH OF            SELECTED PATH                                                                 820 FOR PATH = 0 TO PREF.PATH                                                 830 PREF.SCORE(PATH) = 0                                                      840 FOR X = 0 TO PREF.PATH                                                    850 IF RECEIVED.CODES(PREFERRED.PATH(SELECTED.PATH, PATH)) >                  RECEIVED.CODES(PREFERRED.PATH(SELECTED.PATH, X)) THEN PREF.SCORE(PATH) =      PREF.SCORE(PATH) + 1                                                          860 NEXT X                                                                    870 PREF.PATH$(PREF.SCORE(PATH)) = PREFERRED.PATH$(SELECTED.PATH, PATH)       880 NEXT PATH                                                                 890 `---DISPLAY RESULTS-------------------------------                        900 CLS                                                                       910 PRINT #1, "": PRINT #1, USING " PN  OK S(ok) NOK S(nok) CS(ok)            PP(##) RC "; SELECTED.PATH                                                    920 FOR PATH = 0 TO LAST.PATH                                                 930 IF PREFERRED.PATH$(SELECTED.PATH, PATH) = "" THEN                         PREFERRED.PATH$(SELECTED.PATH, PATH) = STRING $((LOG(LAST.PATH) / LOG(        10) +                                                                         .5), ""                                                                       940 IF PREFERRED.PATH$(SELECTED.PATH, PATH) = "" THEN                         PREFERRED.PATH $(SELECTED.PATH, PATH) = STRING $((LOG(LAST.PATH) /            LOG(10) +                                                                     950 IF PATH = SELECTED.PATH THEN MARK$(PATH) = "*": ELSE MARK$(PATH) = "      960 PRINT #1, USING "&##  +###s. ## +###s. ##  ##  &  ###";                   MARK$(PATH); PATH; TIME(OK, PATH); SCORE(OK, PATH); TIME(NOK, PATH);          SCORE(NOK, PATH); CORRECTED.SCORE(OK, PATH);                                  PREFERRED.PATH$(SELECTED.PATH, PATH); RECEIVED.CODES(PATH)                    970 NEXT PATH                                                                 980 PRINT #1, "": PRINT #1, USING "PP(##) RC  S(rc)"; SELECTED.PATH           990 FOR PATH = 0 TO PREF.PATH                                                 1000 PRINT #1, USING " &  ##  ##"; PREFERRED.PATH$(SELECTED.PATH,             PATH); RECEIVED.CODES(PREFERRED.PATH(SELECTED.PATH, PATH));                   PREF.SCORE(PATH)                                                              1010 NEXT PATH                                                                1020 PRINT #1, : PRINT #1, USING "CPO(1): ##"; SELECTED.PATH;                 1030 FOR PATH = PREF.PATH TO 0 STEP -1                                        1040 PRINT #1, ; VAL(PREF.PATH$(PATH));                                       1050 NEXT PATH                                                                1060 FOR X = LAST.PATH TO 0 STEP -1                                           1070 FORPATH = 0 TO LAST.PATH                                                 1080 IF PATH = SELECTED.PATH THEN NOPRINT(PATH) = 1                           1090 FOR Y = PREF.PATH TO 0 STEP -1                                           1100 IF VAL(PREF.PATH$(Y)) = PATH THEN NOPRINT(PATH) = 1                      1110 NEXT Y                                                                   1120 IF (SCORE(OK, PATH) = X) AND (NOPRINT(PATH) = O) THEN PRINT #1,;         PATH;                                                                         1130 NEXT PATH                                                                1140 NEXT X                                                                   1150 PRINT #1,                                                                1160 IF TIME(NOK, SELECTED.PATH) >-300 THEN PRINT #1, "CPO(2): x "; :         ELSE                                                                          PRINT #1, "CPO(2): "; STR$(SELECTED.PATH); "";                                1170 FOR PATH = PREF.PATH TO 0 STEP -1                                        1180 IF TIME(NOK, VAL(PREF.PATH$(PATH))) >-300 THEN PRINT #1, "x "; :         ELSE                                                                          PRINT #1, ; PREF.PATH$(PATH); "";                                             1190 NEXT PATH                                                                1200 FOR X = LAST.PATH TO 0 STEP -1                                           1210 FOR PATH = 0 TO LAST.PATH                                                1220 IF PATH = SELECTED.PATH THEN NOPRINT(PATH) = 1                           1230 FOR Y = PREF.PATH TO 0 STEP -1                                           1240 IF VAL(PREF.PATH$(Y)) = PATH THEN NOPRINT(PATH) = 1                      1250 NEXT Y                                                                   1260 IF (TIME(NOK, PATH) > -300) AND (SCORE(OK, PATH) = X) AND                (NOPRINT(PATH) = 0) THEN PRINT #1, "x"; : ELSE IF (SCORE(OK, PATH) = X)       AND                                                                           (NOPRINT(PATH) = 0) THEN PRINT #1, ; STR$(PATH);                              1270 NEXT PATH                                                                1280 NEXT X                                                                   1290 PRINT #1,                                                                1300 PRINT #1, "CPO(3): "; : IF TIME(NOK, SELECTED.PATH) < -300 THEN          PRINT #1,                                                                     STR$(SELECTED.PATH); "";                                                      131 FORPATH = PREF.PATH TO 0 STEP -1                                          1320 IF TIME(NOK, VAL(PREF.PATH$(PATH))) < -300 THEN PRINT #1,;               PREF.PATH$(PATH); "";                                                         1330 NEXT PATH                                                                1340 FOR X = LAST.PATH TO 0 STEP -1                                           1350 FOR PATH = 0 TO LAST.PATH                                                1360 IF PATH = SELECTED.PATH THEN NOPRINT(PATH) = 1                           1370 FOR Y = PREF.PATH TO 0 STEP -1                                           1380 IF VAL(PREF.PATH$(Y)) = PATH THEN NOPRINT(PATH) = 1                      1390 NEXT Y                                                                   1400 IF (TIME(NOK, PATH) < -300) AND (SCORE(OK, PATH) = X) AND                (NOPRINT(PATH) = 0) THEN PRINT #1, ; STR$(PATH);                              1410 NEXT PATH                                                                1420 NEXT X                                                                   1430 CLOSE                                                                    __________________________________________________________________________

We claim:
 1. A telecommunication method for a system in which a localuser is capable of being called from at least one central station by wayof transmission paths selected by a selection device, said methodcomprising the steps of:calculating a selection order prior to a call tothe local user, in accordance with probabilities per transmission paththat the call will be answered by the local user, by processing at leastone selection table comprising a designation of reply probabilities pertransmission path, and thereafter selecting the transmission paths inconformity with the calculated selection order, until the call isanswered by the local users, wherein said at least one selection tablecomprises at least: (i) a first selection table formed as a result of aprocessing of registrations, in at least one said central station, ofcode signals received from the local user, including points in timethereof, and (ii) a second selection table including points in time atwhich code signals were transmitted to the local user by way of thetransmission paths, but which, as shown by a lack of receipt of a codesignal from the local user in response thereto, remained unanswered bythe local user.
 2. A telecommunication method according to claim 1,wherein said first selection table includes points in time at which codesignals were received from the local user by way of the transmissionpaths.
 3. A telecommunication method according to claim 2, furthercomprising the step of assigning a score to each transmission path inaccordance with said points in time at which code signals were receivedfrom the local user such that a more recent point in time at which acode signal was received from the local user receives a higher score. 4.A telecommunication method according to claim 1, further comprising thestep of assigning a score to each transmission path in accordance withsaid points in time at which code signals remained unanswered by thelocal user such that a more recent point in time at which a code signalremained unanswered by the local user receives a lower score.
 5. Atelecommunication method according to claim 1, wherein said at least oneselection table may comprise a third selection table having adesignation of a number of code signals received from the local user pertransmission path.
 6. A telecommunication method according to claim 5,wherein the number of code signals received from the local user islowered as a function of time.
 7. A telecommunication method accordingto claim 5, further comprising the step of assigning a score to eachtransmission path in accordance with said number of code signalsreceived from the local user such that a greater number of code signalsreceived from the local user receives a higher score.
 8. Atelecommunication method according to claim 1, wherein said at least oneselection table may comprise a fourth selection table having preferredtransmission paths per transmission path.
 9. A telecommunication methodaccording to claim 8, wherein said fourth selection is formed based onone of:(a) cartographic information, and (b) other geographicinformation.
 10. A telecommunication method according to any one ofclaim 3, 4, 7 or 8, further comprising the step of using, as a firsttransmission path for calling the local user, a selected one of thetransmission paths which has a highest score.
 11. A telecommunicationmethod according to claim 10, further comprising the step of, in anevent of failure of a call by way of the first transmission path,repeating said call by way of at least one of the preferred transmissionpaths in said fourth selection table which is associated with said firsttransmission path.
 12. A telecommunication method according to claim 11,further comprising the step of repeating the call by way of one of thepreferred transmission paths in said fourth selection table in asequence corresponding to the scores of the preferred transmission pathswith respect to one of the first, second and third selection tables. 13.A telecommunication method according to claim 11, further comprising thestep of, in an event of the call by way of the preferred transmissionpaths in said fourth selection table also being unsuccessful, repeatingthe call by way of a remaining one of the transmission paths inaccordance with a sequence corresponding to the scores of the remainingtransmission paths with respect to one of the first, second and thirdselection tables.
 14. A telecommunication method according to claim10,further comprising the step of ruling out respective ones of thetransmission paths for which the first selection table includes pointsin time at which code signals were received from the local user shorterthan a specific threshold value time.